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I'm trying to pick up a little bit of Galois theory to see its applications to classical geometry and constructions of regular $n$-gons, the regular $17$-gon to be exact. This problem comes from Hartshorne's Classical Geometry and appears in Section 29 on Gauss's construction of a regular $17$-gon.

Let $\zeta=\cos(2\pi/7)+i\sin(2\pi/7)$ and let $\alpha=\zeta+\zeta^{-1}$.

(a) Find the minimal polynomial for $\alpha$ over $\mathbb{Q}$.

(b) Show that $\mathbb{Q}(\zeta)$ contains a unique subfield of $E$ of degree $2$ over $\mathbb{Q}$. Find an integer $d$ for which $E=\mathbb{Q}(\sqrt{d})$.

For $(a)$, I noticed that $\alpha=\zeta+\zeta^{-1}=\zeta+\zeta^6=2\cos(2\pi/7)$. Also, $$ \alpha^2=\zeta^2+2+\zeta^5 $$ $$ \alpha^3=\zeta^3+\zeta^4+3\alpha $$ and so $$ \alpha^3+\alpha^2+\alpha=\zeta^6+\zeta^5+\zeta^4+\zeta^3+\zeta^2+\zeta+2+3\alpha $$ and then $\alpha^3+\alpha^2-2\alpha=1$ since $\zeta$ is a root of $\Phi_7$, the seventh cyclotomic polynomial. So I believe the minimal polynomial of $\alpha$ is $x^3+x^2-2x-1$, which is irreducible by the rational roots test.

However for part (b), I'm coming up blank. I couldn't find an expression for $\zeta$ in terms of square roots, so I'm not really sure what $\mathbb{Q}(\zeta)$ looks like, nor can I hazard a guess as to what $d$ may be. I figured if $\zeta=\sqrt{10+2\sqrt{5}}$ or something, I could guess $d=5$ and attempt to work with that. Unfortunately, I'm not too keen with algebra, so my question is, how would one go about part (b)? Thanks.

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up vote 12 down vote accepted

Since $\zeta$ satisfies $x^6 + x^5 + x^4 + x^3 + x^2 + x + 1$, which is irreducible over $\mathbb{Q}$ (e.g., do Eisenstein to $p(x+1)$), then $\mathbb{Q}(\zeta)$ is of degree $6$ over $\mathbb{Q}$. Moreover, since the "other" roots of this polynomial are $\zeta^i$, $i=2,\ldots,6$, then $\mathbb{Q}(\zeta)$ is the splitting field of this polynomial, and the Galois group is completely determined by what it does to $\zeta$. Since $[\mathbb{Q}(\zeta)\colon\mathbb{Q}]=6$, the Galois group is either isomorphic to $S_3$ or to $\mathbb{Z}_6$ (cyclic of order $6$). Either way, there is a unique subgroup of order $3$, whose fixed field is a subextension of degree $2$ over $\mathbb{Q}$. This is what you are looking for.

Some corrections below to using a basis for the extension.

Now, the elements of $\mathbb{Q}(\zeta)$ are of the form $$a + b\zeta + c\zeta^2 + d\zeta^3 + e\zeta^4 + f\zeta^5 +g\zeta^6,\qquad a,b,c,d,e,f,g\in\mathbb{Q},$$ which can be rewritten to be in the form $$a + b\zeta + c\zeta^2 + d\zeta^3 + e\zeta^4 + f\zeta^5$$ (for other rationals) using the relation $\zeta^6 = -1-\zeta-\zeta^2-\zeta^3-\zeta^4-\zeta^5$; the expression is then unique.

In fact, the Galois group is isomorphic to $\mathbb{Z}_6$: the map that sends $\zeta$ to $\zeta^3$ has order $6$ and determines an automorphism. The subgroup of order $3$ is generated by the square of this element, which sends $\zeta$ to $\zeta^2$. The subextension of degree $2$ is the fixed field of this automorphism. This automorphism maps $$\begin{align*} a+b\zeta+c\zeta^2 + d\zeta^3 + e\zeta^4 + f\zeta^5 &\longmapsto a + b\zeta^2 + c\zeta^4 + d\zeta^6+ e\zeta + f\zeta^3\\ &= a + b\zeta^2 + c\zeta^4 + d(-1-\zeta-\zeta^2-\zeta^3-\zeta^4-\zeta^5) + e\zeta + f\zeta^3\\ &= (a-d)+ (e-d)\zeta + (b-d)\zeta^2 + (f-d)\zeta^3 + (c-d)\zeta^4 -d\zeta^5. \end{align*}$$

If the element is fixed by this map, we must have $d=0$ (so $a=a-d$), hence $f=0$, and $b=c=e$.

So an element is fixed by this map if and only if it is of the form $$a + b(\zeta + \zeta^2 + \zeta^4),\qquad a,b\in\mathbb{Q}.$$

Any such element which is not in $\mathbb{Q}$ will generate the extension. So taking $$\zeta+\zeta^2+\zeta^4$$ will do. Now you can do the same sort of thing you did in part (a). If you square $\zeta+\zeta^2+\zeta^4$, you get $$\begin{align*} (\zeta+\zeta^2+\zeta^4)^2 &= \zeta^2 + \zeta^4 + \zeta + 2\zeta^3 + 2\zeta^5 + 2\zeta^6\\ &= -1 + \zeta^3+\zeta^5+\zeta^6\\ &= -2-(\zeta+\zeta^2+\zeta^4); \end{align*}$$ thus you have that $\beta=\zeta+\zeta^2+\zeta^4$ satisfies the polynomial $$x^2 + x + 2.$$ This polynomial is irreducible over $\mathbb{Q}$, and its discriminant is $-7$. So the splitting field is given by $\mathbb{Q}(\sqrt{-7})$. This is the field you want.

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Thank you Arturo. I follow your calculations, but I have a small theory question about the beginning. Forgive me if it is obvious, I'm getting this Galois theory from an appendix in Hartshorne where there are no proofs. How exactly do you see that the Galois group is isomorphic to $\mathbb{Z}_6$? The map sending $\zeta\mapsto\zeta^3$ has order $6$ with $\zeta$ as a generator, like $\zeta\mapsto\zeta^3\mapsto\zeta^2\mapsto\zeta^6\mapsto\zeta^4\mapsto\zeta^5$? Is that how you see the Galois group is cyclic? And then is the subextension degree $2$ since $6/3=2$, and the degree of field extensions – yunone Apr 17 '11 at 23:04
is multiplicative? As in if $F\subseteq E\subseteq G$, then $\deg(G/F)=\deg(G/E)\deg(E/F)$? – yunone Apr 17 '11 at 23:05
@yunone: Since the degree of the extension is 6, the Galois group has six elements. Since the map that sends $\zeta$ to $\zeta^3$ (and $\zeta^3$ to $\zeta^2$, etc) has order $6$, and is an element of the Galois group. That means that consider this map and all its powers you already get six things in the Galois group. Since the Galois group has six things in it altogether, then these maps are all the things in the Galois group. Since they are all powers of the same element, that tells you that the Galois group is cyclic. (cont) – Arturo Magidin Apr 17 '11 at 23:07
@yunone: Your equation makes no sense, because $G$ is the Galois group and $F$ is a field. Instead: if $G$ is the Galois group of $K$ over $F$, then the subgroups of $H$ correspond to the subfields of the extension by letting $H$ corresponds to $E$, $F\subseteq E\subseteq K$, where $E$ is the fixed set of $H$; and $L$, $F\subseteq L\subseteq K$ corresponds to the subgroup $M$ of all things that fix $L$. If $H$ corresponds to $E$, then $|H| = \deg(K/E)$, and therefore the index of $H$ in $G$ equals $\deg(E/F)$, because the degree is multiplicative (and $|G|=|H|\times\mathrm{index}(H)$) – Arturo Magidin Apr 17 '11 at 23:10
Oh ok, thanks! I was confused over what exactly the Galois group was. Thank you for the extra explanation. – yunone Apr 17 '11 at 23:14

This is just a remark to the effect that Arturo's answer, although mostly correct, contains a few inaccuracies. Specifically, the field $\mathbf{Q}(\zeta)$ is of degree $6$, and so one needs to be careful writing elements in terms of the (seven) elements $\zeta^i$ for $i = 0$ to $6$. It would probably be better to write elements of $\mathbf{Q}(\zeta)$ as $a + b \zeta + c \zeta^2 + d \zeta^3 + e \zeta^4 + f \zeta^5$ and replace all occurrences of $\zeta^6$ by $$\zeta^6 = -1 - \zeta - \zeta^2 -\zeta^3 - \zeta^4 - \zeta^5.$$

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Hrmph. Good point; I thought of that, got distracted, and never went back to fix it. – Arturo Magidin Apr 18 '11 at 1:47

In this case $d = -7$. In general, if $p \equiv 1 \mod{4}$ then $\mathbb{Q}(\sqrt{p}) \subseteq \mathbb{Q}(\zeta_p)$ and if $p \equiv 3 \mod{4}$ then $\mathbb{Q}(\sqrt{-p}) \subseteq \mathbb{Q}(\zeta_p)$. There are several proofs of this fact, but the only elementary one that comes to mind involves Gauss sums: Let $g = \sum_{a=1}^6 \chi(a) \zeta_7^a$ where $\chi(a) = \pm 1$ depending on whether or not $a$ is a quadratic residue mod $7$. You can show that $g\bar{g} = 7$ but that $\bar{g} = -g$. So $g^2 = -7$. So $g = \pm \sqrt{-7}$. So $\mathbb{Q}(\zeta_7)$ contains $\mathbb{Q}(\sqrt{-7})$.

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